5. The length of similar components produced by a company are approximated by a normal distribution model with a mean of 5 cm and a standard deviation of 0.02 cm. If a component is chosen at random a) what is the probability that the length of this component is between 4.98 and 5.02 cm

the probability that  he length of this component is between 4.98 and 5.02 cm is 0.682 (68.2%)

Step-by-step explanation:

Since the random variable X= length of component chosen at random , is normally distributed, we can define the following standardized normal variable Z:

Z= (X- μ)/σ

where μ= mean of X  , σ= standard deviation of X

for a length between 4.98 cm and 5.02 cm , then

Z₁= (X₁- μ)/σ =  (4.98 cm - 5 cm)/0.02 cm = -1

Z₂= (X₂- μ)/σ = (5.02 cm - 5 cm)/0.02 cm = 1

therefore the probability that the length is between 4.98 cm and 5.02 cm is

P( 4.98 cm ≤X≤5.02 cm)=P( -1 ≤Z≤ 1) = P(Z≤1) - P(Z≤-1)

from standard normal distribution tables we find that

P( 4.98 cm ≤X≤5.02 cm) = P(Z≤1) - P(Z≤-1) = 0.841 - 0.159 = 0.682 (68.2%)

therefore the probability that  he length of this component is between 4.98 and 5.02 cm is 0.682 (68.2%)

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: